Turkish Mathematical Olympiad

Since $\angle RPD = \angle RAD = \angle A'AC = \angle A'PC = \angle DPC$, $P, R, C$ are collinear. Then $\angle PRD = \angle PAD = \angle PAB = \angle PCB$ implies that $DR \parallel BC$. Similarly, $SE \parallel BC$, and consequently, $SE \parallel DR$ and $\frac{VD}{DA} = \frac{SR}{RA}$.

Let $l$ be tangent line through $A$ to the circumcircle of $ABC$, and let $T$ and $U$ be the points where $l$ intersects the lines $DS$ and $SE$, respectively. Also let $V$ be the point of intersection of $AB$ and $SE$.

We have $\angle UAE = \angle UAC = \angle B$, $\angle AEU = \angle AVE + \angle EAV = \angle B + \angle A$, and $\angle UAS = \angle UAE + \angle EAS = \angle AVS + \angle SAV = \angle ASU$. From these we conclude that $US = UA$ and then $\frac{US}{UE} = \frac{UA}{UE} = \frac{\sin \angle C}{\sin \angle B} = \frac{SV}{ES}$ where we used the law of sines in the triangle $AEU$ and the fact that $AS$ is the angle bisector of the triangle $VAE$.

$$\frac{AT}{TU} \cdot \frac{US}{SV} \cdot \frac{VD}{DA} = 1 \text{ by Menelaus' Theorem for the triangle } AUV \text{ and the line } DS.$$
Substituting the last two ratios from above we obtain
$$\frac{AT}{TU} \cdot \frac{UE}{ES} \cdot \frac{SR}{RA} = 1$$
Applying Menelaus' Theorem for the triangle $AUS$ we conclude that the points $R, E, T$ are collinear.